Let n be the number obtained on rolling a fair die.

Question

Let $n$ be the number obtained on rolling a fair die. If the probability that the system
$$x-ny+z=6,\quad x-(n-2)y+(n+1)z=8,\quad (n-1)y+z=1$$
has a unique solution is $\dfrac{k}{6}$, then the sum of $k$ and all possible values of $n$ is:

Accepted Answer

The system has a unique solution when the coefficient determinant is non-zero:
$$\begin{vmatrix}1 & -n & 1\ 1 & -(n-2) & n+1\ 0 & n-1 & 1\end{vmatrix}=0 \iff n=1,\,2,\text{ or } n=-1\ (\text{rejected as }n\ge1).$$

So the determinant vanishes only at $n=1,2$; a unique solution exists for $n=3,4,5,6$ --- 4 out of 6 outcomes.

$P(\text{unique solution})=\dfrac{4}{6}\Rightarrow k=4$.

Required sum $=k+(3+4+5+6)=4+18=22$.

Solutions · 1